Specific Heat, Molar Specific Heat and Cp-Cv Relation
The First Law and Energy Conservation
The First Law of Thermodynamics states that energy added to a System as heat minus the work done by the system equals the change in its Internal Energy.
$$\Delta U = Q - W$$
Energy conservation statement for thermodynamic processes.
$$\Delta U$$=Change in internal energy(J)
$$Q$$=Net heat added to the system(J)
$$W$$=Work done BY the system(J)
Adiabatic process ($Q = 0$)
→$$\Delta U = -W$$ (work done at expense of internal energy)
Isothermal process ($\Delta U = 0$)
→$$Q = W$$ (all heat converts to work)
Isochoric (constant volume, $W = 0$)
→$$\Delta U = Q$$ (all heat stored internally)
Sign Convention: Heat entering the system is positive ($+Q$); work done by the system is positive ($+W$). Work done ON the system makes $W$ negative.
State vs Path: Internal Energy depends only on the current state (State Function), while $Q$ and $W$ are Path Function variables.
Dimensional Check: All three quantities ($\Delta U$, $Q$, $W$) have units of joules — verify unit consistency before substituting.
Isothermal and Adiabatic Processes
An isothermal process occurs at constant temperature. For an ideal gas, Internal Energy depends only on temperature, so $\Delta U = 0$ and all heat converts to work.
$$P_1 V_1 = P_2 V_2$$
Boyle's Law holds during isothermal changes — pressure and volume are inversely proportional.
$$P_1, V_1$$=Initial pressure and volume(Pa, m³)
$$P_2, V_2$$=Final pressure and volume(Pa, m³)
Volume doubles ($V_2 = 2V_1$)
→Pressure halves ($P_2 = P_1/2$)
Slow Process: Isothermal changes must occur slowly to allow continuous heat exchange and maintain constant temperature.
First Law Reduction: Since $\Delta U = 0$, the first law gives $Q = W$ — heat absorbed equals work done by the gas.
PV Diagram: The curve is a rectangular hyperbola called an isotherm.
An adiabatic process is one where no heat enters or leaves the system ($Q = 0$). Work done by the gas comes entirely from its internal energy, causing the temperature to change.
$$PV^\gamma = \text{constant}$$
The adiabatic condition — pressure drops faster than in isothermal expansion because temperature also decreases.
$$\gamma$$=Ratio of molar specific heats $C_p / C_v$(dimensionless)
First Law Reduction: With $Q = 0$, the first law gives $W = -\Delta U$. Expansion cools the gas; compression heats it.
Rapid Process: Adiabatic changes occur rapidly (no time for heat exchange) or in insulated systems.
Real Examples: Burst tyre (rapid expansion → cooling), bicycle pump (rapid compression → heating), cloud formation (rising air expands adiabatically).
Steeper Curve: On a PV diagram the adiabat falls more steeply than the isotherm because both pressure and temperature decrease during expansion.
Reversible and Irreversible Processes
A reversible process can be retraced in exactly reverse order without producing any change in the surroundings. All real processes are irreversible to some degree.
Reversible Examples: Slow compression of a gas in a cylinder, slow evaporation and condensation of a substance.
Irreversible Examples: Explosions, friction, rapid heat transfer, free expansion — any process involving dissipation.
Cycle: A succession of events that returns the system to its initial state. A reversible cycle has all reversible steps.
Key Distinction: Reversible processes are idealizations that set the theoretical upper bound on efficiency. Real processes always fall short.
Heat Engines and Thermal Efficiency
A heat engine converts thermal energy into mechanical work by absorbing heat $Q_H$ from a high-temperature source and rejecting waste heat $Q_L$ to a cold Heat Reservoir.
$$\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}$$
Efficiency compares the useful work output to the total heat energy paid for.
$$\eta$$=Thermal efficiency(dimensionless (0 to 1))
$$W$$=Net work output(J)
$$Q_H$$=Heat absorbed from hot reservoir(J)
$$Q_L$$=Heat rejected to cold reservoir(J)
Cyclic Process: The engine must return to its initial state each cycle, so $\Delta U_{\text{cycle}} = 0$ and $W = Q_H - Q_L$.
Energy Balance: By the First Law, $Q_H = W + Q_L$ — input heat splits into useful work and waste heat.
Real-world Limits: Petrol engines ≈ 25-30%, diesel engines ≈ 35-40%, steam turbines ≈ 35-46%.
The Second Law of Thermodynamics
The Second Law restricts the direction of natural processes: heat flows spontaneously only from hot to cold, and no engine can convert heat entirely into work.
Kelvin Statement: It is impossible to convert heat extracted from a single reservoir entirely into work without leaving any change in the working system.
Practical Implication: Two reservoirs at different temperatures are essential for converting heat into work. Oceans contain vast thermal energy, but without a colder reservoir, none can be extracted as work.
Perpetual Motion: The second law forbids perpetual motion machines of the second kind (100% heat-to-work conversion).
Refrigerator: A heat engine running in reverse — it requires external work to transfer heat from cold to hot surroundings.
The Carnot Cycle
The Carnot Cycle is an ideal reversible cycle consisting of two Isothermal and two Adiabatic steps. No engine can be more efficient than a Carnot engine operating between the same two temperatures.
$$\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$$
The theoretical maximum efficiency depends only on reservoir temperatures in Kelvin.
$$T_L$$=Cold reservoir temperature(K (must be Kelvin))
$$T_H$$=Hot reservoir temperature(K (must be Kelvin))
$T_L = 0$ K
→$\eta = 100\%$ (impossible — absolute zero cannot be reached)
$T_L = T_H$
→$\eta = 0\%$ (no temperature gradient, no work possible)
$T_H$ doubled while $T_L$ fixed
→$\eta$ increases but never reaches 100%
Carnot's Theorem: All Carnot engines between the same two temperatures have the same efficiency, regardless of the working substance.
Proportionality Shortcut: $\eta = (T_H - T_L)/T_H$. If $T_H = 2T_L$, then $\eta = 50\%$. If $T_H = 3T_L$, then $\eta = 66.7\%$.
Raising Efficiency: Since $T_L$ is usually near room temperature, the only practical way to increase efficiency is to raise $T_H$.
Four Steps of the Carnot Cycle (PV Diagram)
1
A→B: Isothermal expansion at $T_H$ — gas absorbs $Q_H$, does work, volume increases
2
B→C: Adiabatic expansion — temperature drops from $T_H$ to $T_L$, no heat exchange
3
C→D: Isothermal compression at $T_L$ — gas rejects $Q_L$, work done on gas
4
D→A: Adiabatic compression — temperature rises from $T_L$ back to $T_H$
Petrol and Diesel Engines
Practical heat engines like petrol and diesel engines operate on four-stroke cycles based on the Carnot principle, but with irreversible processes that reduce their efficiency.
Petrol Engine (4 strokes): (1) Intake — fuel-air mixture drawn in, (2) Compression — adiabatic compression, (3) Power — spark ignites mixture, rapid expansion drives piston, (4) Exhaust — waste gases expelled.
Diesel Engine Difference: No spark plug needed. Air compressed to very high temperature, then diesel sprayed in and ignites on contact (compression ignition).
Efficiency Comparison: Diesel engines (35-40%) are more efficient than petrol engines (25-30%) because of higher compression ratios.
Multi-cylinder Design: Cars typically use 4+ cylinders on the same crankshaft, firing in sequence for smooth power delivery.
Entropy: The Measure of Disorder
Entropy ($S$) is a thermodynamic State Function that measures the unavailability of a system's energy to do work — often interpreted as molecular disorder.
$$\Delta S = \frac{\Delta Q}{T}$$
Entropy change equals heat transferred divided by absolute temperature (for a reversible process).
$$\Delta S$$=Change in entropy(J/K (or J K⁻¹))
$$\Delta Q$$=Heat transferred (positive if added, negative if removed)(J)
$$T$$=Absolute temperature at which transfer occurs(K)
Reversible process
→$\Delta S_{\text{total}} = 0$ (entropy of universe unchanged)
Irreversible process
→$\Delta S_{\text{total}} > 0$ (entropy of universe always increases)
Sign Convention: $\Delta S > 0$ when heat is added to the system; $\Delta S < 0$ when heat is removed.
Net Entropy of Heat Transfer: When $Q$ flows from $T_H$ to $T_L$: $\Delta S_{\text{net}} = Q/T_L - Q/T_H > 0$ (always positive since $T_L < T_H$).
Proportionality Insight: Same $Q$ at low $T$ creates MORE disorder than at high $T$ — warming ice by 1 J changes entropy more than warming steam by 1 J.
Phase Changes: Melting and boiling increase entropy (more disorder); freezing and condensation decrease local entropy.
Degradation of Energy: Increasing entropy means energy degrades from useful, ordered forms to random thermal motion.
The entropy statement of the Second Law: any natural process proceeds in the direction that increases the total entropy of the system plus its environment.
Arrow of Time: Entropy increase defines the direction of natural processes — you can scramble an egg but never unscramble it.
Mixing Example: When hot and cold water are mixed, the warm result cannot spontaneously separate. Energy is conserved but is no longer available for work.
Free Expansion: Gas expanding into a vacuum increases entropy — molecules gain greater randomness of position.
Environmental Impact: Thermal pollution is an inevitable consequence of the Second Law. All energy processes increase the entropy of the environment, and even small temperature changes can disrupt ecological balance.
Thermodynamic Temperature Scale
The Carnot cycle provides a temperature scale independent of any material's properties, using the ratio of heat transfers at two temperatures.
$$T = 273.16 \frac{Q}{Q_{\text{tr}}}$$
Unknown temperature determined by comparing heat exchange to that at the triple point of water.
$$T$$=Unknown temperature(K)
$$Q$$=Heat absorbed/rejected at temperature $T$(J)
$$Q_{\text{tr}}$$=Heat absorbed/rejected at triple point of water(J)
Triple Point: The unique state where ice, liquid water, and water vapour coexist in equilibrium at exactly 273.16 K.
Material-Independent: Unlike mercury or alcohol thermometers, this scale does not depend on any substance's expansion properties.
Kelvin Unit: 1 K is defined as 1/273.16 of the thermodynamic temperature of the triple point of water.
Application Strategy
Worked Examples
Problem Statement
A gas enclosed in a container with a piston (area 0.10 m², pressure 8000 N/m²) is heated slowly. The piston moves up 4.0 cm while 42 J of heat is transferred. Find the change in internal energy.
First Law of ThermodynamicsWork Done by Expansion
Strategic Toolkit
$$\Delta U = Q - W, \quad W = P \Delta V$$
Step 1
Calculate Work
Work done by the gas during expansion at constant pressure: $W = P \times A \times \Delta y$.
$$W = 8000 \times 0.10 \times 0.04 = 32\text{ J}$$
Step 2
Apply First Law
Substitute $Q = 42$ J and $W = 32$ J.
$$\Delta U = 42 - 32 = 10\text{ J}$$
Step 3
Interpret
Of the 42 J of heat, 32 J went to expansion work and only 10 J was stored as internal energy.
Problem Statement
A steam power plant takes steam from a boiler at 427°C and exhausts into a reservoir at 77°C. What is the maximum possible efficiency?
Carnot EfficiencyTemperature Unit Conversion
Strategic Toolkit
$$\eta = 1 - \frac{T_L}{T_H}$$
Step 1
Convert to Kelvin
Thermodynamic temperatures must be in Kelvin: $T(\text{K}) = T(°\text{C}) + 273$.
$$T_H = 427 + 273 = 700\text{ K}$$
$$T_L = 77 + 273 = 350\text{ K}$$
Step 2
Calculate Efficiency
Apply the Carnot formula.
$$\eta = 1 - \frac{350}{700} = 1 - 0.5 = 0.5$$
Step 3
Final Answer
Express as a percentage.
$$\eta = 50\%$$
Problem Statement
Calculate the entropy change when 1.0 kg of ice at 0°C melts into water at 0°C. Latent heat of fusion $L_f = 3.36 \times 10^5$ J/kg.
Entropy ChangePhase Transitions
Strategic Toolkit
$$\Delta S = \frac{\Delta Q}{T} = \frac{m L_f}{T}$$
Step 1
Convert Temperature
The melting occurs at 0°C, which must be converted to Kelvin.
$$T = 0 + 273 = 273\text{ K}$$
Step 2
Calculate Heat
Total heat absorbed during melting.
$$\Delta Q = m L_f = 1.0 \times 3.36 \times 10^5 = 3.36 \times 10^5\text{ J}$$
Step 3
Find Entropy Change
Divide heat by absolute temperature.
$$\Delta S = \frac{3.36 \times 10^5}{273} = 1.23 \times 10^3\text{ J/K}$$
Step 4
Interpret
Positive $\Delta S$ confirms that melting increases disorder — water molecules have more freedom of motion than ice molecules.
Problem Statement
A reversible engine works between two temperatures whose difference is 100°C. It absorbs 746 J from the source and rejects 546 J to the sink. Calculate the source and sink temperatures.
Carnot EfficiencyEnergy BalanceSimultaneous Equations
Strategic Toolkit
$$\eta = \frac{W}{Q_H} = \frac{T_H - T_L}{T_H}$$
Step 1
Calculate Efficiency from Heat
Use the heat values to find efficiency.
$$\eta = \frac{Q_H - Q_L}{Q_H} = \frac{746 - 546}{746} = \frac{200}{746} = 0.268$$
Step 2
Set Up Temperature Equation
Carnot efficiency also equals $(T_H - T_L)/T_H$. We know $T_H - T_L = 100$ K.
$$0.268 = \frac{100}{T_H} \implies T_H = \frac{100}{0.268} = 373\text{ K} = 100°\text{C}$$
Step 3
Find Sink Temperature
Subtract the temperature difference.
$$T_L = 373 - 100 = 273\text{ K} = 0°\text{C}$$
Memorize formulas
First Law of Thermodynamics
Energy balance: Heat in minus work out equals storage change.
Formula
$$\Delta U = Q - W$$
Adiabatic Condition
Pressure-volume relation when no heat is exchanged ($Q = 0$).
Formula
$$PV^\gamma = \text{constant}$$
Thermal Efficiency
What you get (Work) vs what you pay for (Heat Input).
Formula
$$\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}$$
Carnot Efficiency Limit
The absolute ceiling for efficiency using absolute temperatures (Kelvin).
Formula
$$\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$$
Entropy Change
Heat transferred divided by absolute temperature at which transfer occurs.
Formula
$$\Delta S = \frac{\Delta Q}{T}$$
Review
1
A heat engine absorbs 500 J of heat from a hot reservoir and performs 150 J of work. What is the efficiency of the engine and how much heat is exhausted?
Ans
Efficiency 30%, Exhaust 350 J
Review
Tap to reveal
Compute the efficiency from work output
For a heat engine, efficiency is the fraction of absorbed heat converted into work. Use the given work and heat input directly.
$$\eta = \frac{W}{Q_H} = \frac{150}{500} = 0.30 = 30\%$$
Find the rejected heat
The remainder of the absorbed heat is exhausted to the cold reservoir.
$$Q_L = Q_H - W = 500 - 150 = 350\ \text{J}$$
2
A Carnot engine operates between 600 K and 300 K. If the hot reservoir temperature is increased by 100 K, what is the new maximum efficiency?
Ans
57.1%
Review
Tap to reveal
Update the hot-reservoir temperature
The hot source rises by 100 K, so the Carnot engine now operates between 700 K and 300 K.
$$T_H = 600 + 100 = 700\ \text{K}$$
Apply the Carnot efficiency formula
Maximum efficiency depends only on the two reservoir temperatures.
$$\eta_{\max} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{700} = 0.571 \approx 57.1\%$$
3
An ideal gas at 600 K is compressed adiabatically and 500 J of work is done ON the gas. What is the final temperature direction and the change in internal energy?
Ans
Temperature increases, $\Delta U = +500$ J
Review
Tap to reveal
Use the adiabatic condition
No heat enters or leaves the gas, so the First Law reduces to a work-energy balance.
$$Q = 0$$
Convert work done on the gas into the sign convention
If 500 J is done on the gas, the work done by the gas is negative.
$$W = -500\ \text{J}$$
Find the internal-energy change and infer the temperature trend
Substitute into $\Delta U = Q - W$. A positive increase in internal energy for an ideal gas means the temperature rises.
$$\Delta U = 0 - (-500) = +500\ \text{J},\quad T\ \text{increases}$$
4
An engineer claims his engine operating between 327°C and 27°C has an efficiency of 52%. Is this claim valid?
Ans
No, the Carnot limit for these temperatures is 50%
Review
Tap to reveal
Convert the temperatures to kelvin
Carnot efficiency must use absolute temperatures, so convert both reservoir temperatures from Celsius to kelvin.
$$T_H = 327 + 273 = 600\ \text{K},\quad T_L = 27 + 273 = 300\ \text{K}$$
Check the Carnot limit
The maximum possible efficiency is fixed by the reservoir temperatures.
$$\eta_{\max} = 1 - \frac{300}{600} = 0.50 = 50\%$$
Compare the claim with the upper bound
Any claimed efficiency above 50% violates the Second Law.
$$52\% > 50\% \Rightarrow \text{claim is not valid}$$
5
In an isothermal expansion of an ideal gas, which quantity is zero?
Ans
Change in internal energy ($\Delta U = 0$)
Review
Tap to reveal
Use the temperature dependence of internal energy
For an ideal gas, internal energy is a function of temperature only. If temperature does not change, internal energy does not change either.
$$\Delta U = 0$$
Reject the other choices by the First Law
In an isothermal expansion, the gas still does work and absorbs heat, so neither $W$ nor $Q$ is zero, and entropy increases.
$$Q = W \neq 0,\quad \Delta S > 0$$
6
336 J of energy is required to melt 1 g of ice at 0°C. What is the change in entropy of 30 g of water at 0°C as it is changed to ice by a refrigerator?
Ans
−36.9 J/K
Review
Tap to reveal
Find the heat removed during freezing
Freezing 30 g of water releases the latent heat of fusion for each gram.
$$Q = 30 \times 336 = 10{,}080\ \text{J}$$
Use the entropy-change relation for a reversible temperature
Because heat is being removed from the water, the water's entropy change is negative.
$$\Delta S = -\frac{Q}{T} = -\frac{10{,}080}{273} = -36.9\ \text{J/K}$$
7
A reversible engine absorbs 746 J from the source and rejects 546 J to the sink. If the temperature difference between source and sink is 100°C, what is the source temperature?
Ans
100°C (373 K)
Review
Tap to reveal
Compute the engine efficiency from heat and work
The work output is the difference between absorbed and rejected heat.
$$W = 746 - 546 = 200\ \text{J},\quad \eta = \frac{W}{Q_H} = \frac{200}{746} \approx 0.268$$
Use the reversible-engine temperature relation
For a reversible engine, the efficiency depends only on the reservoir temperatures, and their difference is given as 100 K.
$$\eta = \frac{T_H - T_L}{T_H} = \frac{100}{T_H}$$
Solve for the source temperature
Set the two efficiency expressions equal and isolate $T_H$.
$$0.268 = \frac{100}{T_H} \Rightarrow T_H \approx 373\ \text{K} = 100^\circ\text{C}$$
8
A Carnot engine operating between 500 K and 300 K produces 800 J of work per cycle. How much heat does it absorb from the hot reservoir per cycle?
Ans
2000 J
Review
Tap to reveal
Find the Carnot efficiency
The reservoir temperatures determine the maximum possible efficiency.
$$\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{500} = 0.40$$
Recover the absorbed heat from the work output
Use the efficiency definition $\eta = W/Q_H$ and solve for the input heat.
$$Q_H = \frac{W}{\eta} = \frac{800}{0.40} = 2000\ \text{J}$$
9
Which of the following is an irreversible process?
Ans
A chemical explosion
Review
Tap to reveal
Identify the process that cannot be retraced
Irreversible processes create large dissipative effects and cannot be run backward without leaving changes in the surroundings.
$$\text{chemical explosion} \rightarrow \text{irreversible}$$
Eliminate the quasi-static options
Slow or quasi-static processes are the opposite of an explosion and are treated as approximately reversible.
$$\text{slow compression, slow evaporation, quasi-static expansion} \rightarrow \text{approximately reversible}$$
10
A thermos flask containing milk is shaken rapidly. According to the First Law, what happens to the milk's temperature?
Ans
It increases, because mechanical work is converted to internal energy
Review
Tap to reveal
Treat the thermos as nearly adiabatic
An insulated thermos exchanges very little heat with its surroundings during shaking.
$$Q \approx 0$$
Account for work done on the milk
Rapid shaking transfers mechanical energy into the milk, increasing its internal energy.
$$\Delta U = Q - W \approx -W > 0$$
State the temperature effect
For the milk, a rise in internal energy means a rise in temperature.
$$T\ \text{increases}$$
11
Why can the vast thermal energy of the oceans not be used to run a heat engine?
Ans
There is no reservoir at a sufficiently lower temperature to serve as a heat sink
Review
Tap to reveal
Recall the requirement for a heat engine
A heat engine needs a hot reservoir and a colder sink so that heat can flow and some work can be extracted.
$$T_H > T_L$$
Apply that requirement to the oceans
The ocean is not separated from its environment by a large enough temperature difference to provide a useful sink.
$$\text{no significant } \Delta T \Rightarrow \eta_{\text{Carnot}} \approx 0$$
Question 1 of 11
0:00
1
A heat engine absorbs 500 J of heat from a hot reservoir and performs 150 J of work. What is the efficiency of the engine and how much heat is exhausted?